Absolute parallelism in teleparallel gravity

[karlzr]

I am reading Hayashi's paper titled "http://prd.aps.org/abstract/PRD/v19/i12/p3524_1" " (Phys. Rev. D 19, 3524, 1979).

In Riemann-Cartan geometry, one can choose the Weitzenbock connection rather than the well-known Levi-Civita connection, so that Riemann curvature tensor vanishes while the torsion tensor arises. In this case, the gravitational effect is attributed to the torsion tensor exclusively. This theory is dubbed teleparallel gravity since absolute parallel vectors can be defined in this geometry, because of the vanishing of curvature tensor, as the authors stated.

So I don't quite follow the authors' logic. Does vanishing curvature necessarily guarantee absolute parallelism?
In general(or in general relativity), no absolute parallelism exists because vectors parallel transported along a closed infinesimal parallelogram to the starting position will be different from the original one. The difference is dependent on both curvature and torsion tensor. Quoting Sean Carroll's textbook:
[\nabla_\mu,\nabla_\nu]V^\rho=R^\rho{}_{\sigma\mu\nu}V^\sigma-T^\lambda{}_{\mu\nu}\nabla_\lambda V^\rho
If absolute parallelism exists, I believe the second term on the right-hand side of the above equation should vanishes simultaneously. So any comments on this issue?

 

[HallsofIvy]

Yes, if the curvature of a Riemann space is 0 at every point, then it is Euclidean.

 

[karlzr]

Yes, if the curvature of a Riemann space is 0 at every point, then it is Euclidean.

What about the torsion? Besides, in teleparallel gravity, local Lorentz symmetry is broken.
I still can't see why it should be Euclidean or Minkowski ?

 

[yenchin]

What about the torsion? Besides, in teleparallel gravity, local Lorentz symmetry is broken.
I still can't see why it should be Euclidean or Minkowski ?

I think of it this way (might be wrong?): the spacetime is originally not flat if you are using Levi-Civita connection. What teleparallel theories do is to choose an orthonormal frame at any point, define a new connection (the Weitzenbock connection) and parallel translate with respect to that connection, then by construction, spacetime as viewed in that frame is flat, which means you can parallel transport vectors to any point without ambiguity (path independent) and thus obtain a parallel vector field on the manifold (globally if the manifold is parallelizable, indeed all 3-manifolds and all (3+1)-manifolds are parallelizable). As you say, local Lorentz symmetry is broken, i.e. teleparallel theories are a theory of global Lorentz symmetry only (a priori there is nothing wrong with that). This is not a problem for TEGR (which is obtained by writing the Einstein-Hilbert action in terms of contractions of torsion tensors instead of Ricci scalar) as the field equation does not uniquely determine a parallelization, so you could have chosen another tetrad to begin with and the result would be the same. This is however not true in other teleparallel theories with is not equivalent to GR, including the recently popular f(T) theories. See e.g. discussions in http://arxiv.org/abs/1012.4039. I am still trying to understand much of these.

In other words the spacetime thus has *two* structures - there are two connections - Levi-Civita and Weitzenbock. In TEGR (teleparallel equivalent of GR), you can re-write every formulas of GR involving Riemann curvature in terms of the torsions. Note for example that free particles still follow the same geodesics of GR which is determined by Levi-Civita connection, not the geodesic equations determined by the Weitzenbock connection. In teleparallel language, there is a contortion term that is the source of gravity (this is equivalent to geodesics in GR) which deviates particles from following Weitzenbock geodesics curves. In other words, you can think of TEGR as GR written in terms of another connection. Of course there are many versions of teleparallel theories, and TEGR is but one of them. Also see e.g. http://arxiv.org/abs/gr-qc/9706070.

 

[karlzr]

I think of it this way (might be wrong?): the spacetime is originally not flat if you are using Levi-Civita connection. What teleparallel theories do is to choose an orthonormal frame at any point, define a new connection (the Weitzenbock connection) and parallel translate with respect to that connection, then by construction, spacetime as viewed in that frame is flat, which means you can parallel transport vectors to any point without ambiguity (path independent) and thus obtain a parallel vector field on the manifold (globally if the manifold is parallelizable, indeed all 3-manifolds and all (3+1)-manifolds are parallelizable). ……

Let me try to interpret your ideas. What teleparallel gravity intends to do is to find a specific connection with respect to which path-indenpendent parallel transportation can be defined. In this regard, we obtain the Weitzenbock connection. Meanwhile, it is just a coincidence that the curvature tensor defined with this connection vanishes.
What confuses me in this issue is exactly the relation between absolute parallelism and vanishing curvature tensor.

 

[yenchin]

Let me try to interpret your ideas. What teleparallel gravity intends to do is to find a specific connection with respect to which path-indenpendent parallel transportation can be defined. In this regard, we obtain the Weitzenbock connection. Meanwhile, it is just a coincidence that the curvature tensor defined with this connection vanishes.
What confuses me in this issue is exactly the relation between absolute parallelism and vanishing curvature tensor.

Yes, but I wouldn't call it a coincidence: since the connection is specifically designed to have path-independent parallel transport, it would have vanishing Riemannian curvature.

Anyway just to add on, in teleparallel geometry, there is a preferred orthonormal frame in which the connection one-forms vanish in the frame. But there is another fine print: another condition that we use is that the tetrad (like the metric) is covariant constant (metric compatible), and that the tetrad is related to the metric in a certain way. But one could contemplate an even more general theory in which this is not the case [e.g. http://arxiv.org/abs/gr-qc/9809049, it also has some discussions on teleparallel theories].

From a more mathematical point of view, John Baez talks about teleparallel stuffs here: http://math.ucr.edu/home/baez/week176.html in item 13.

 

[Mentz114]

Let me try to interpret your ideas. What teleparallel gravity intends to do is to find a specific connection with respect to which path-indenpendent parallel transportation can be defined. In this regard, we obtain the Weitzenbock connection. Meanwhile, it is just a coincidence that the curvature tensor defined with this connection vanishes.
What confuses me in this issue is exactly the relation between absolute parallelism and vanishing curvature tensor.

Have look at equation (15) and the following lines, especially the top of page 4 in the overview paper arXiv:gr-qc/0011087v1.

 

[karlzr]

since the connection is specifically designed to have path-independent parallel transport, it would have vanishing Riemannian curvature.

Maybe I didn't put it clearly. Actually, the quoted sentence is exactly what confuses me. Why path-independent leads to vanishing curvature?

I know this result can be derived through rigorous calculations, using Weitzenbock connection. But I believe this point could be illustrated in a more direct manner, which may have something to do with Carroll's equation listed in the original post. Perhpas it's quite trivial to you guys, but not to me. So more details would be sincerely appreciated.

 

[Mentz114]

Maybe I didn't put it clearly. Actually, the quoted sentence is exactly what confuses me. Why path-independent leads to vanishing curvature?

I know this result can be derived through rigorous calculations, using Weitzenbock connection. But I believe this point could be illustrated in a more direct manner, which may have something to do with Carroll's equation listed in the original post. Perhpas it's quite trivial to you guys, but not to me. So more details would be sincerely appreciated.

What makes vectors rotate ( i.e. become unparallel) is curvature. In the absence of curvature all displaced vectors remain parallel to themselves. Absolute parallelism. OK ?

The second term in your equation above represents a closure deficit of translated points, not a rotation, so it can be non-zero without affecting parallelism.

Torsion measures the noncommutativity of displacements of points in analogy to the curvature tensor which measures the noncommutativity of displacements of vectors.

Gronwald & Hehl, in arXiv:gr-qc/9602013.

vectors parallel transported along a closed infinesimal parallelogram to the starting position will be different from the original one.

But they will be parallel - that's what parallel transport means. There will be a closure deficit if torsion is non-zero.